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d1i1m1o1n [39]
2 years ago
6

Calculate how many grams of the product form when 16.7 g of calcium metal completely reacts. Assume that there is more than enou

gh of the chlorine gas.
Ca(s) + Cl2(g) → CaCl2(s)
Chemistry
1 answer:
swat322 years ago
4 0

39.96 g product form when 16.7 g of calcium metal completely reacts.

<h3>What is the stoichiometric process?</h3>

Stoichiometry is a section of chemistry that involves using relationships between reactants and/or products in a chemical reaction to determine desired quantitative data.

Equation:

Ca(s) + Cl_2(g) → CaCl_2(s)

In this case, for the undergoing reaction, we can compute the grams of the formed calcium chloride by noticing the 1:1 molar ratio between calcium and it (stoichiometric coefficients) and using their molar mass of 40 g/mol and 111 g/mol by using the following stoichiometric process:

m_{ca_C_l_2}= 16.7 g Ca x \frac{1 mol \;of \;Ca}{40g Ca} x \frac{1 mol \;of \;CaCl_2}{1 mol \;Ca} x \frac{111g of \;CaCl_2}{1 mol \;CaCl_2}

m_{ca_C_l_2} = 39.96 g

Hence, 39.96 g product form when 16.7 g of calcium metal completely reacts.

Learn more about the stoichiometric process here:

brainly.com/question/15047541

#SPJ1

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3 years ago
Fermentation is a complex chemical process of wine making in which glucose is converted into ethanol and carbon dioxide: C6H12O6
attashe74 [19]

Answer:

Mass of ethanol obtained is 260.682 grams.

Volume of the ethanol obtained is 0.3304 L.

Explanation:

C_6H_{12}O_6\rightarrow 2C_2H_5OH + 2CO_2

Moles of glucose = \frac{510.6 g}{180 g/mol}=2.833 mol

According to the reaction, 1 mol of gluocse gives 2 moles of ethanol.

Then 2.833 moles of glucose will give :

\frac{2}{1}\times 2.8333 mol=5.667 mol

Mass of 5.667 moles of an ethanol :

5.667 mol × 46 g/mol = 260.682 g

Volume of ethanol = V

Density of the ethanol = 0.789 g/mL

Density=\frac{Mass}{Volume}

0.789 g/mL=\frac{260.682 g}{V}

V=\frac{260.682 g}{0.789 g/mL}=330.40 mL = 0.3304 L

Mass of ethanol obtained is 260.682 grams.

Volume of the ethanol obtained is 0.3304 L.

5 0
3 years ago
Some people think that wind power should be used for all the electricity in the United States. Which argument could be used to d
Marianna [84]
There might not be ample enough wind to power all the electricity in the United States and in some places there might not be wind at all so people wouldn't be able to have electricity.
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3 years ago
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Which of the following can serve as evidence to support the claim that human consumption of water impacts earths system
Brilliant_brown [7]

Ouestion: Which of the following can serve as evidence to support the claim that human consumption of water impacts earths system?

Answer & Explanation: Typically as human populations and per-capita consumption of natural resources increase, so do the negative impacts on Earth unless the activities and technologies involved are engineered otherwise. (MS-ESS3-3), (MS-ESS3-4)

Hope this helps and comment down below if you need more information!

Fr0ggyLikeSMELLY

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2 years ago
The British gold sovereign coin is an alloy of gold and copper having a total mass of 7.988 g, and is 22-karat gold.
azamat

Answer:

(a) m_{gold}=7.322g

(b)

V_{gold}=0.379cm^3

V_{copper}=0.122cm^3

(c) \rho _{coin}=15.94g/cm^3

Explanation:

Hello,

(a) In this case, with the given formula we easily compute the mass of gold contained in the sovereign  as shown below:

m_{gold}=\frac{m_{tota}*karats}{24}=\frac{7.988g*22}{24}=7.322g

(b) Now, by knowing the density of gold and copper, 19.32 and 8.94 g/cm³ respectively, we compute each volume, by also knowing that the rest of the coin contains copper:

V_{gold}=\frac{m_{gold}}{\rho_{gold}} =\frac{7.322g}{19.32g/cm^3}=0.379cm^3

m_{copper}=7.988g-7.322g=1.09g\\V_{copper}=\frac{m_{copper}}{\rho_{copper}}=\frac{1.09g}{8.94g/cm^3}  \\\\V_{copper}=0.122cm^3

(c) Finally, the volume is computed by dividing the total mass over the total volume containing both gold and copper:

\rho _{coin}=\frac{m_{total}}{V_{gold}+V_{copper}}=\frac{7.988 g}{0.379cm^3+0.122cm^3}\\  \\\rho _{coin}=15.94g/cm^3

Best regards.

3 0
3 years ago
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